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%I #7 Jun 13 2017 22:36:40
%S 1,1,1,1,8,1,1,232,64,1,1,36968,16192,512,1,1,35593832,21928768,
%T 1047040,4096,1,1,219379963496,178379459392,11424946688,67096576,
%U 32768,1,1,9003699178010216,9288403489672000,748093366229504,5862250172416
%N Triangle P, read by rows, that satisfies [P^8](n,k) = P(n+1,k+1) for n>=k>=0, also [P^(8*m)](n,k) = [P^m](n+1,k+1) for all m, where [P^m](n,k) denotes the element at row n, column k, of the matrix power m of P, with P(0,k)=1 and P(k,k)=1 for all k>=0.
%C Also P(n,k) = partitions of (8^n - 8^(n-k)) into powers of 8 <= 8^(n-k).
%F Let q=8; the g.f. of column k of P^m (ignoring leading zeros) equals: 1 + Sum_{n>=1} (m*q^k)^n/n! * Product_{j=0..n-1} L(q^j*x) where L(x) satisfies: x/(1-x) = Sum_{n>=1} Product_{j=0..n-1} L(q^j*x)/(j+1) and L(x) equals the g.f. of column 0 of the matrix log of P (A111839).
%e Let q=8; the g.f. of column k of matrix power P^m is:
%e 1 + (m*q^k)*L(x) + (m*q^k)^2/2!*L(x)*L(q*x) +
%e (m*q^k)^3/3!*L(x)*L(q*x)*L(q^2*x) +
%e (m*q^k)^4/4!*L(x)*L(q*x)*L(q^2*x)*L(q^3*x) + ...
%e where L(x) satisfies:
%e x/(1-x) = L(x) + L(x)*L(q*x)/2! + L(x)*L(q*x)*L(q^2*x)/3! + ...
%e and L(x) = x - 6/2!*x^2 + 142/3!*x^3 + 31800/4!*x^4 +... (A111839).
%e Thus the g.f. of column 0 of matrix power P^m is:
%e 1 + m*L(x) + m^2/2!*L(x)*L(8*x) + m^3/3!*L(x)*L(8*x)*L(8^2*x) + m^4/4!*L(x)*L(8*x)*L(8^2*x)*L(8^3*x) + ...
%e Triangle P begins:
%e 1;
%e 1,1;
%e 1,8,1;
%e 1,232,64,1;
%e 1,36968,16192,512,1;
%e 1,35593832,21928768,1047040,4096,1;
%e 1,219379963496,178379459392,11424946688,67096576,32768,1; ...
%e where P^8 shifts columns left and up one place:
%e 1;
%e 8,1;
%e 232,64,1;
%e 36968,16192,512,1; ...
%o (PARI) P(n,k,q=8)=local(A=Mat(1),B);if(n<k || k<0,0, for(m=1,n+1,B=matrix(m,m);for(i=1,m, for(j=1,i, if(j==i || j==1,B[i,j]=1,B[i,j]=(A^q)[i-1,j-1]);));A=B); return(A[n+1,k+1]))
%Y Cf. A111836 (column 1), A111837 (row sums), A111838 (matrix log); triangles: A110503 (q=-1), A078121 (q=2), A078122 (q=3), A078536 (q=4), A111820 (q=5), A111825 (q=6), A111830 (q=7).
%K nonn,tabl
%O 0,5
%A _Gottfried Helms_ and _Paul D. Hanna_, Aug 22 2005
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